Let $A$ denote the set of all integers except $i$. Since $G_A$ is transitive, it follows that there exists an element $\sigma \in G$ that fixes $i$ and sends any fixed $j \ne i$ to some $k \ne i$. From this one quickly deduces that $G$ is $2$-transitive, and hence primitive.
Choose a prime $p$ such that $n-3 \ge p > n/2$ (by a modification of Chebyshev's proof of Bertrand's postulate, this is true whenever $n \ge 8$). Let $B = \{1,2, \ldots, p\}$. Since the stabilizer $G_B \subset G$ of $B$ acts transitively on $B$, by the orbit-stabilizer theorem, $G_B$ (and hence $G$) has an element of order $p$. Since $n-p < p$, it follows that $G$ contains a $p$-cycle. But a well known theorem of Jordan says that any primitive permutation group containing a $p$ cycle for $p \le n-3$ is the symmetric or alternating group.
This leaves $n \le 7$. Note that since $G_A$ acts transitively for any $A$, it follows that $|G|$ is divisible by $|A|$.
For $n = 1$, $2$, and $3$, $G$ has to be $S_n$ (easy). For $n = 4$ and $5$, $|G|$ must be divisible by $12$ and $60$ respectively. So $G = A_n$ or $S_n$ in these cases. Yet both $A_n$ and $S_n$ satisfy the defining condition when $n \ge 4$. For $n = 7$, $|G|$ must be divisible by $420$. The only primitive subgroups of $S_7$ with order divisible by $420$ are $A_7$ and $S_7$. This leaves $n = 6$, where $|G|$ is divisible by $60$. It follows that $G$ could possibly be $A_5$, $S_5$, $A_6$, or $S_6$. The latter two groups work. The group $S_5$ with its natural embedding (stabilizing one point) does not work, because it does not act transitively on the full set $\{1,2,\ldots,6\}$. The same remark applies to the standard embedding of $A_5$. It suffices, then, to check the exotic actions of $A_5$ and $S_5$. Here is a "computation free" argument.
This exotic action of $S_5$ on $6$ points is realized by the action of $\mathrm{PGL}_2(\mathbf{F}_5)$ on the projective line over $\mathbf{F}_5$. The action of $\mathrm{PGL}_2(\mathbf{F}_5)$ is sharply $3$-transitive, and the action of $A_5 = \mathrm{PSL}_2(\mathbf{F}_5)$ is $2$-transitive. It follows that both groups satisfy the conditions for $|A| = 1$ and $|A| = 2$. The first group satisfies the condition for $|A| = 3$ since it is $3$-transitive. Suppose that $|A| = 5$. Then we may assume by transitivity that $A = \mathbf{F}_5 = \mathbf{P}^1(\mathbf{F}_5) \setminus \{\infty\}$. Both groups act transitively on this set via the element $z \mapsto z + 1$. Finally, let $|A| = 4$. Since both groups are transitive, $2$-transitive, we may assume that $A = \mathbf{F}^{\times}_5 = \mathbf{P}^1(\mathbf{F}_5) \setminus \{0,\infty\}$. The subgroup of elements of $\mathrm{PGL}_2(\mathbf{F}_5)$ which fix $0$ and $\infty$ are of the form $z \mapsto a z$ for $a \in \mathbf{F}^{\times}_5$, which acts transitively. The set of elements of $\mathrm{PGL}_2(\mathbf{F}_5)$ which send $0$ to $\infty$ and $\infty$ to $0$ are of the form $z \mapsto a/z$ for $a \in \mathbf{F}^{\times}_5$. Yet the corresponding elements in $\mathrm{PSL}_2(\mathbf{F}_5)$ are of the form $z \mapsto a^2 z$ and $z \mapsto a^2/z$, which does not act transitively since it only sends quadratic residues to quadratic residues. Thus the complete list of such groups is:
(i) $G = S_n$ for $n \le 3$.
(ii) $G = A_n$, $S_n$ for $n \ge 4$.
(iii) $G = \mathrm{PGL}_2(\mathbf{F}_5) = S_5$ when $n = 6$.
